A particle is moving along the $x-$axis with its coordinate with the time '$t$' given be $\mathrm{x}(\mathrm{t})=10+8 \mathrm{t}-3 \mathrm{t}^{2} .$ Another particle is moving the $y-$axis with its coordinate as a function of time given by $\mathrm{y}(\mathrm{t})=5-8 \mathrm{t}^{3} .$ At $\mathrm{t}=1\; \mathrm{s},$ the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{\mathrm{v}} .$ Then $\mathrm{v}$ (in $\mathrm{m} / \mathrm{s})$ is

  • [JEE MAIN 2020]
  • A

    $441$

  • B

    $600$

  • C

    $580$

  • D

    $484$

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  • [AIPMT 1989]

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